Answer to number 8 please

Physics

2 answers:

jeyben [28]3 years ago

7 0

Acceleration = (change in speed) / (time for the change)

(Note: That's all the Physics there is to this problem.
The rest is all arithmetic.)
=======================================

change in speed = (40 - 20) miles/hour = -20 miles/hour
time for the change = 10 minutes

Acceleration = (-20 miles/hour) / (10 minutes) =

-2 miles/hour per minute .

That's a perfectly good and technically correct expression for acceleration.
But obviously the units might make some people dizzy. So let's try to
clean it up a little.

Notice that 10 minutes is 1/6 of an hour.
So we could write the acceleration as

Acceleration = (-20 miles/hour) / (1/6 hour) = -120 miles/hour per hour =

-120 miles/hour² .

You could convert this into any units you like. It's really not a physics problem
any more, it's just an exercise in converting units.

eduard3 years ago

3 0

First segment: The airplane is descending slowly, with time and distance at a 3:1 ratio. It is probably approaching the airport

Second segment: The airplane is maintaining its altitude. It is probably waiting for the other planes to clear away

Third segment:The airplane descending rapidly, with time and distance at a 1:2 ratio. It is probably landing

Sorry, my bad.
To find acceleration, we use v2-v1 /t
40mph - 20mph / 10 min
20mph/10min
We can convert this to 20/6 miles per 10 minutes, and cancel out the 10 min to 20/6 miles, which is 10/3 miles

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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b

erma4kov [3.2K]

Answer:

$h = 269.6 m$

Explanation:

Pressure at the bottom of the building and at the top of the building must be related as

$P_{top} = P_{bottom} - \rho g h$

$P_{top} = 730 mm Hg$

$P_{bottom} = 755 mm Hg$

now we will have

$(755 \times 10^{-3})(13.6 \times 10^3)(9.81) = P_{bottom}$

$P_{bottom} = 1.007 \times 10^5 Pa$

$P_{top} = (730\times 10^{-3})(13.6 \times 10^3)(9.81)$

$P_{top} = 0.974 \times 10^5$

now we have

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$h = 269.6 m$

7 0

3 years ago

a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur

Doss [256]

Refer to the diagram shown below.

Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer: 10.2 m/s at 21.5° downstream.